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1 year ago

5 (a)

Mathematics

2 answers:

jeyben [28]1 year ago

7 0

Answer:

Step-by-step explanation:

We can have a set of equations for this. Let x be the number of 1 pound coins and y be the number of 2 pound coins.

We can set an equation like this:

Since she has twice as many 1 pound than 2 pound coins, we can set this equation:

x=2y

Since the worth of the coins add up to 100:

(1 pound)*x+(2 pounds)*y=100pounds

x+2y=100

We can substitute x=2y

2y+2y=100

y=25

Therefore we have 25 2 pound coins

andrey2020 [161]1 year ago

4 0

Answer:

Step-by-step explanation:

25 2 euro coins*2=50

50 1 euro coins*1=50

50+50=100

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Which ratio is different? 3 : 7 3 to 7 3/7 7 to 3

Masteriza [31]

Answer:

7 to 3 is this answer

Step-by-step explanation:

if you noticed, with all the other ones they all have 3 : 7 in it. So by the process of elimination it's 7 to 3 that is different. I'll tell you why,

3 : 7, that is the ratio we have as an example. That's why it isn't the different one.

3 to 7, you say 3 : 7 as 3 to 7 to it still has the same ratio

3/7, this is how to type the fraction version of the ratio so it makes sense that it is the same one as 3 : 7.

7 to 3, if you haven't noticed yet, all of them has 3 : 7 as a counterpart. They are all the same but this one, this is 7 : 3 or 7 to 3. So with logic, this is the different ratio.

5 0

2 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$